Here is a sketch of a more intuitive geometric approach, though the technical details get fairly involved. The basic idea is to triangulate $U$ and use simplicial homology, so that a nontrivial class in $H_n(U)$ would be forced to be a linear combination of fundamental classes of connected components of $U$ (where the fundamental class of a component is just the sum of all the $n$-simplices in that component), which is not actually possible since $U$ is not compact so this would be an infinite sum of simplices.
As a first step, you prove there exists a triangulation of $U$ by linear simplices in $\mathbb{R}^n$. There are various ways to prove this; for instance, you can take a sequence of increasingly fine cubical meshes on $\mathbb{R}^n$ and use this to write $U$ as a nice union of cubes, and then subdivide those cubes into simplices to get a triangulation.
Now since $U$ is an $n$-manifold, this triangulation has the property that each $(n-1)$-simplex is a face of exactly two $n$-simplices (if an $(n-1)$-simplex was a face of a different number of $n$-simplices, then $U$ would not be locally homeomorphic to $\mathbb{R}^n$ at a point in the interior of that $(n-1)$-simplex). This means that if $\alpha$ is a simplicial $n$-cycle with respect to this triangulation that has a nonzero coefficient on a certain $n$-simplex $\sigma$, then $\alpha$ is forced to also have a nonzero coefficient on each $n$-simplex that shares a face with $\sigma$, since that is the only way to cancel out each boundary face of $\sigma$ to make $\partial\alpha$ equal to $0$. Now the idea is that every other $n$-simplex in the same connected component as $\sigma$ can be reached by repeatedly taking $n$-simplices that share boundary faces in this way, and so every $n$-simplex in the same connected component as $\sigma$ must appear in $\alpha$. But since no connected component of $U$ is compact, this would mean infinitely many $n$-simplices must appear in $\alpha$, which is a contradiction. Thus no $n$-simplex $\sigma$ can have a nonzero coefficient in $\alpha$, so $\alpha=0$.
It remains to prove rigorously that every $n$-simplex in the same connected component as $\sigma$ can be reached from $\sigma$ by repeatedly taking $n$-simplices that share boundary faces. This follows from a connectedness argument using the fact that simplices of dimension less than $n-1$ can be "ignored" for the purposes of connectedness. Let $V$ be the connected component of $U$ containing $\sigma$, let $V_0$ be the $(n-2)$-skeleton of $V$, and let $S$ be the set of $n$-simplices and $(n-1)$-simplices of the triangulation which are contained in $V$. Define an equivalence relation $\sim$ on $S$ by saying two simplices are equivalent if one can be reached from the other by repeatedly passing between $n$-simplices and their boundary faces. If $A\subseteq S$ is an equivalence class with respect to $\sim$, let $V(A)\subseteq V$ be the union of the interiors of all the simplices of $A$. Note that $V(A)$ is connected and open in $V\setminus V_0$. Moreover, these sets $V(A)$ form a partition of $V\setminus V_0$ into open sets as $A$ ranges over all the equivalence classes in $S$. So to conclude that there is only one such equivalence class (so every $n$-simplex in $V$ can be reached from $\sigma$ by repeatedly passing through shared boundary faces), it suffices to show that $V\setminus V_0$ is connected.
To prove that $V\setminus V_0$ is connected, take two points $p,q\in V\setminus V_0$. Since $V$ is a connected open subset of $\mathbb{R}^n$, we can connect $p$ and $q$ by a piecewise linear path in $V$. Now the idea is that you can perturb the sequence of endpoints $(p,x_1,\dots,x_m,q)$ of the linear pieces of this path to avoid $V_0$. There are various ways to formulate this argument; for instance, you can consider the sequence $(x_1,\dots,x_m)$ of intermediate endpoints of the linear pieces as an element of $(\mathbb{R}^n)^m$. Then, for any linear simplex $\sigma\subset\mathbb{R}^n$ of dimension at most $n-2$, the set of such sequences in $(\mathbb{R}^n)^m$ for which the piecewise linear path will intersect $\sigma$ is closed and nowhere dense (this boils down to the fact that given two points in $\mathbb{R}^n$, you can perturb them so that the line segment between them does not pass through $\mathbb{R}^{n-2}$; by ignoring the first $n-2$ coordinates, you can reduce this to the case $n=2$ where it is obvious). Since $V_0$ is a union of countably many such simplices, by the Baire category theorem, the subset of $(\mathbb{R}^n)^m$ consisting of sequences that give piecewise linear paths that avoid $V_0$ is dense. So, we can pick such a sequence that is sufficiently close to $(x_1,\dots,x_m)$ so that the piecewise linear path remains in $V$, and thus we get a path between $p$ and $q$ in $V\setminus V_0$.
